How to calculate protons and electrons in NH3BF3?

1. The problem statement, all variables and given/known data
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?

2. Relevant equations
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?

3. The attempt at a solution
Don't know how to calculate in NH3^+- BF3^-

Staff: Mentor

Not sure what your question is. When the compound (molecule) is neutral, we know number of electrons and number protons are identical. Number of protons in nucleus is what identifies an element, so all is given here.

Not sure what your question is. When the compound (molecule) is neutral, we know the number of electrons and number protons are identical. The number of protons in the nucleus is what identifies an element, so all is given here.

Staff: Mentor

No, I am not going to tell you anything like that. Each time you ask a question and you are pointed to the way of solving the problem, you still ask us to do the calculations for you. We are not here to spoon feed you, it is up to you to try.

NH3 - how many nitrogen atoms? How many hydrogen atoms? How many protons in the nitrogen nucleus? How many protons in the hydrogen nucleus? How many protons in total?

No, I am not going to tell you anything like that. Each time you ask a question and you are pointed to the way of solving the problem, you still ask us to do the calculations for you. We are not here to spoon feed you, it is up to you to try.

NH3 - how many nitrogen atoms? How many hydrogen atoms? How many protons in the nitrogen nucleus? How many protons in the hydrogen nucleus? How many protons in total?

1. one nitrogen atom.
2. three hydrogen atoms.
3. 7 protons in the nitrogen nucleus.
4. three protons in hydrogen nuclei.
5. Total 10 protons. 7 from Nitrogen and 3 from hydrogen.
But I want to know that how many protons in NH3+ ion?

If I remove 1 Hydrogen atom from ammonia, then it would be NH3, not NH3+ because N has a lone pair of electrons in NH3. As far as I know, to become NH3+, N ha to donate its paired electrons. So how is it possible?

Looks like you are confused about everything at once. First things first: NH3+ is an ion. How do you make ions? (hint: have you heard about ionization energy? Electron affinity?)

Yes, I have heard ionizing energy. The energy required to remove an electron from a neutral atom and the electron affinity is the energy released when an electron is added to a neutral atom. Earlier, I made a mistake by writing NH3. If I remove one H from NH3, it would be NH2. I am sorry. So now I got NH2. Now what to do to get NH3+?

It doesn't have to be an atom - actually you can ionize any molecule in the gaseous state this way.

(Which hints at how to make NH3+ from NH3.)

Now I understand that If I want to get NH3+, I have to remove electrons from NH3. But still, I have a doubt here. How many electrons I have to remove to get NH3+ from NH3. May it be one, two, three or four electrons. I am confused.

Staff: Mentor

When it comes to ions on the molecular level all charges are expressed in electron (or proton - they are identical, just differ in the sign) charges. Thus - compared with the neutral atoms - Cl- has a single excess electron, while Na+ is missing one electron.

When it comes to ions on the molecular level all charges are expressed in electron (or proton - they are identical, just differ in the sign) charges. Thus - compared with the neutral atoms - Cl- has a single excess electron, while Na+ is missing one electron.

I understand what you say but what about NH3BH3 where N has N+ and B has B-. As you said above, ''When it comes to ions on the molecular level all charges are expressed in electron (or proton - they are identical, just differ in the sign) charges. Thus - compared with the neutral atoms - Cl- has a single excess electron, while Na+ is missing one electron'' If I draw the Lewis structure of this molecule NH3, here N has 5 valence electrons and H has 1 valence electron so N makes 3 covalent bonds with 3 H and still it has a lone pair of electrons. So N full fills its octet. On the other hand, If we look at the Lewis structure BF3, here B has 3 valence electrons and F has 7 valence electrons. As a result, B makes three covalent bonds with F and B does not full fill its octet. It needs two more electrons to fills its octet. Now N gives the electron pair to B, making N+ and B- ( a coordinate covalent bond formed between NH3 and BF3).
Now my question is that as you said above, ''Cl- has a single excess electron, while Na+ is missing one electron.'' similarly, can I take ''B- has a single excess electron, while ''N+ is missing one electron? Could you explain it, please?

Staff: Mentor

Now N gives the electron pair to B, making N+ and B- ( a coordinate covalent bond formed between NH3 and BF3).

These are not [itex]N^+[/itex] and [itex]B^-[/itex], but [itex]N^\oplus[/itex] and [itex]B^\ominus[/itex] - atoms bearing partial charges, meaning fraction of an electron (exactly how much depends on many factors and is best determined experimentally).

Just + and just - mean a single charge (expressed in the electron charges).

These are not [itex]N^+[/itex] and ]itex]B^-[/itex], but [itex]N^\oplus[/itex] and [itex]B^\ominus[/itex] - atoms bearing partial charges, meaning fraction of an electron (exactly how much depends on many factors and is best determined experimentally).

Just + and just - mean a single charge (expressed in the electron charges).

I understand what you say a little bit. Could you help me to understand the difference in detail you gave above, please?

Staff: Mentor

Can't help you understand not knowing what you don't understand.

Think about it this way: electron spends some time close to one atom, some time close to other atom. If it stays on one side a bit longer, if we will try to measure the charge we will see a bit of excess of the electron charge there. That means a slight negative charge. At the same time exactly the same charge will be missing on the other side, so we will see a small positive charge there.

Think about it this way: electron spends some time close to one atom, some time close to other atom. If it stays on one side a bit longer, if we will try to measure the charge we will see a bit of excess of the electron charge there. That means a slight negative charge. At the same time exactly the same charge will be missing on the other side, so we will see a small positive charge there.

NH3 + HCl = NH4Cl
Ammonium ions, NH4+, are formed by the transfer of a hydrogen ion from the hydrogen chloride to the lone pair of electrons on the ammonia molecule.When the ammonium ion, NH4+, is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen's nucleus is transferred from the chlorine to the nitrogen. The hydrogen's electron is left behind on the chlorine to form a negative chloride ion. and if we count the electrons NH4+, the total electrons should be 5 from N and 4 from H, 5 + 4 = 9 but due to ''+ ve,'' one electron is missing here. and in ''Cl-'' if we count the electrons, it should be 7 electrons but due to a "- ve" we find a single excess electron here. In this case, the concept is clear to me. but in the case below I still have a doubt.
Similarly like above, in NH3BF3 molecule, Is NH3+ missing one electron and BF3- has a single excess electron? this is my simple question.

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Staff: Mentor

the total electrons should be 5 from N and 4 from H, 5 + 4 = 9 but due to ''+ ve,'' one electron is missing here

Not sure what you are doing here. ±ve is (in the context of molecules and dipoles present on them) usually used to describe partial charges, not full charges. Electron is missing not because of "+ve", but because neutral ammonia accepted a single proton into its lone pair, no electrons moved together with this proton.

Not sure what you are doing here. ±ve is (in the context of molecules and dipoles present on them) usually used to describe partial charges, not full charges. Electron is missing not because of "+ve", but because neutral ammonia accepted a single proton into its lone pair, no electrons moved together with this proton.

You said above that 'neutral ammonia accepted a single proton into its lone pair,'
Could you tell me from which atom neutral ammonia accepts a single proton?

Staff: Mentor

Single proton is a hydrogen nucleus.

Note: NH3 is ammonia, NH4+ is ammonium - that is, ammonia plus a proton. No such thing as NH3+, unless we are talking about some exotic conditions. NH3BF3 doesn't contain NH3+, all we can say is that there is some small positive charge around nitrogen atom (that's what the [itex]N^\oplus[/itex] means) that makes the molecule a dipole. I feel like you are confusing all these things at once.